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Mademuasel [1]
3 years ago
13

Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it.

Mathematics
1 answer:
grandymaker [24]3 years ago
7 0
(2x+1)^{\cot x}=\exp\left(\ln(2x+1)^{\cot x}\right)=\exp\left(\cot x\ln(2x+1)\right)=\exp\left(\dfrac{\ln(2x+1)}{\tan x}\right)

where \exp(x)\equiv e^x.

By continuity of e^x, you have

\displaystyle\lim_{x\to0^+}\exp\left(\dfrac{\ln(2x+1)}{\tan x}\right)=\exp\left(\lim_{x\to0^+}\dfrac{\ln(2x+1)}{\tan x}\right)

As x\to0^+ in the numerator, you approach \ln1=0; in the denominator, you approach \tan0=0. So you have an indeterminate form \dfrac00. Provided the limit indeed exists, L'Hopital's rule can be used.

\displaystyle\exp\left(\lim_{x\to0^+}\dfrac{\ln(2x+1)}{\tan x}\right)=\exp\left(\lim_{x\to0^+}\dfrac{\frac2{2x+1}}{\sec^2x}\right)

Now the numerator approaches \dfrac21=2, while the denominator approaches \sec^20=1, suggesting the limit above is 2. This means

\displaystyle\lim_{x\to0^+}(2x+1)^{\cot x}=\exp(2)=e^2
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